Aclay ball with a mass of .045 kg is attached to a long string to make a pendulum. the ball is pulled back so that the string is horizontal to the ground, and is then released. at the bottom of the ball’s path is another clay ball that has a mass of .075 kg and is at rest. the two balls collide elastically so that they follow the path of the first ball beyond the point of collision.
(a) what must the speed of the first ball be just before the collision so that the combined balls rise to a height of .08 m above the point of collision?
(b) how high must the first ball be raised for it to have this speed at the bottom of its path?

(a) speed=v₁ =3.34 m/s

(b) height=0.569 m

Explanation:

Given Data

Clay ball mass=0.045 kg

Other clay ball mass=0.075 kg

To Find

(a) Speed

(b) Height

Solution

Let

m₁ = mass of the first ball 0.045 kg

m₂ = mass of the two balls together, 0.045 kg + 0.075 kg = 0.120 kg

v₁= velocity of the first ball at the point of collision (to be determined)

v₂ = velocity of the two balls immediately after the collision

d₁ = the height in cm the first ball is dropped from (to be determined)

d₂ = the maximum height the two balls rise to, 8.0 cm

g = acceleration due to gravity 9.80 m/sec²

The kinetic energy of the the balls after the collision is equal to the potential energy at the top of their swing, so

(1/2) × (m₂) × (v₂)² = (m₂) × g × (d₂)

(1/2) × (v₂)² = g × (d₂)

v₂ = √[2×g ×(d₂)]

from the conservation of momentum, we know

(m₁)× (v₁) = (m₂)× (v₂)

(m₁)× (v₁) = (m₂) ×√[2×g ×(d₂)]

(v₁) = [(m₂)/(m₁)]×√[2×g ×(d₂)]

v₁ = (O.120 kg/0.045 kg) ×√(2×9.80 m/s² ×0.08 m)

v₁ =3.34 m/s

As we have find the velocity,Now we can calculate the height the first ball (1/2)× (m₁) × (v₁)² = (m₁)× g× (d₁)

(1/2)× (v₁)² = g×(d₁)

(d₁) = {(1/2) (v1)²} / g

d₁={(1/2) (3.34)²} /9.8

d₁=0.569 m